The sign of f "(x) is the same as the sign of a. The graph of f which is called a parabola will be concave up if a is positive and concave down if a is negative.

Question 2:

Find all intervals on which function f given by

f(x) = sin x

is concave up.

Solution to Question 2:

The first and second derivatives of function f are given by

f '(x) = cos x

f "(x) = - sinx

We now study the sign of f "(x) by solving the inequality

- sin x > 0

Multiply both sides by -1 to obtain

sin x < 0

Let us study sin x < 0 over the interval [0, 2pi] which represents one period of sin x. sin x is negative for x in the interval [pi , 2 pi]. Hence sin x is concave up on the interval (pi , 2 pi). We now extend this result over the domain of f which is (-infinity , +infinity) and say that sin x is concave up on all intervals of the form (pi + 2Pi k , 2pi + 2pi k) where k is an integer taking the values 0, -1, +1, -2, +2,...

Question 3:

The graph of f '(x) is shown below for x in the closed interval [a , g]. On which interval (s) is function f both decreasing and concave down? Find any points of inflection.

Solution to Question 3:

The sign of f '(x) determines the interval of increase / decrease of a function f and the sign of f "(x) dtermines the concavity. We shall divide the whole interval [a , g] into smaller intervals and use a table to study the sign of f ' and the sign of f ".

The sign of f ' is given by the graphs: parts of the graph below the x axis correspond to f '(x) negative and parts above the x axis correspond to f '(x) positive. See table below.

An interval over which f ' increases correspond to f "(x) positive and an interval over which f ' decreases correspond to f "(x) negative. See table below.

The table above shows that f is decreasing and concave down over the intervals (0 , d) and (e , g).

The points of inflection occur when there is a change in concavity. Hence (c , f(c)) , (d , f(d)) , (e , f(e)) and (f , f(g)) are all points of inflection.

Question 4:

Determine all inflection points of function f defined by

f(x) = 4 x 4 - x 3 + 2

Solution to Question 4:

In order to determine the points of inflection of function f, we need to calculate the second derivative f " and study its sign. This gives the concavity of the graph of f and therefore any points of inflection.

f '(x) = 16 x 3 - 3 x 2

f "(x) = 48 x 2 - 6 x

= 6x (8x - 1)

The table below shows the signs of 6x and 8x - 1 and that of f " which is the product of 6x and 8x - 1. Also the concavity is shown. The points of inflection are located where there is a change in concavity. Hence the points

(0 , f(0)) = (0 , 2)

and (1/8 , f(1/8)) = (1/8 , 2047/1024)

are points of inflection.

Question 5:

Determine the inflection point of

f(x) = - x 3 + 3 x 2 + 1

then use it to determine the inflection point of

g(x) = - (x - 2) 3 + 3 (x - 2) 2 + 1

.

Solution to Question 5:

The first and second derivatives of f are given by

f '(x) = - 3 x 2 + 6 x

f "(x) = -6 x + 6

Any change of sign of f " gives an inflection point. The sign of f " changes at x = 1. Hence the inflection point of f is at (1 , f(1)) = (1 , 3).

A close examination of g(x) shows that g(x) = f(x - 2) which means that the graph of g is the graph of f shifted horizontally 2 units to the right. But if the graph of a function is shifted 2 units to the right all points, including any inflection point, on the graph of f are shifted 2 units to the right so that the inflection point of g is at (1+2 , 3) = (3 , 3).